OUR OWN HIGH SCHOOL, Al WARQA’A, DUBAIplaywithurfuture.synthasite.com/resources/chem/Reh....

OUR OWN HIGH SCHOOL, Al WARQA’A, DUBAI REHEARSAL EXAMINATION-1, DEC-2009

Name: ______________________________________ Grade: XII Sec:____

Subject : CHEMISTRY M Marks: 70 Date: 03 - 01- 2010 Time: 3 hours

General Instructions:1. This question paper contains 5 pages.2. All questions are compulsory.3. Question numbers 1 to 8 are very short answer questions each of one mark. 4. Questions 9 to 18 are short answer questions of two marks each. 5. Question numbers 19 to 27 are short answer questions of 3 marks each. 6. Questions 28 to 30 are long answer type questions of 5 marks each.7. Use log tables if necessary. Use of calculators is not allowed.8. Reading time is fifteen minutes.

1. Why is the platinum electrode coated with finely divided platinum in the hydrogen electrode?

[1]

2. Why are zeolites good shape – selective catalysts? [1]

3. What is the role of CaF2 in the extraction of aluminium? [1]

4. Write balanced ionic equation for the action of acidified potassium dichromate on tin(II) chloride solution.

[1]

5. Why does the conductivity of a solution decrease with dilution? [1]

6. Write the IUPAC name of the following compound: OH │ CCH2CHCH3

│ │ CH3 OCH3

[1]

7. State the important functional differences between DNA and RNA. [1]

8. Write the IUPAC name of the following compound: OHCC6H4CHO-p

[1]

9. Which of these two, CdCl2 or NaCl will produce Schottky defect if added to AgCl ? Explain.

[2]

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10. A and B react together to give Z. A + B → Z

Experiments were carried out to determine the order of the reaction with respect to each reagent. In experiment 1, the concentration of A was kept constant but that of B was varied. In experiment 2, the concentration of B was kept constant but that of A was varied.The results were plotted on a graph:

a) From the graph deduce the order of reaction with respect to each reactant.b) Write the rate equation for the reaction.

[2]

11. Given below are the electronic configurations of two members of the lanthanoid series:

a) 4f1 5d1 6s2

b) 4f7 5d0 6s2

Determine their atomic numbers. What oxidation states will possibly be exhibited by these elements? ORIdentify the first row transition metal ions which have outer electronic configurations of 3d3 and 3d6 and describe their oxidation states.

[2]

12. Write one chemical equation each to represent :a) Finkelstein reactionb) Fittig reaction

[2]

13. Assign appropriate reasons for each of the following observations:a) H2SO4 cannot be used along with KI in the conversion of an alcohol into an alkyl iodide.b) CH2Cl undergoes SN2 reaction faster than Cl .

[2]

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14. Complete the following reaction equations:a) C6H5N2Cl + CH3CH2OH

b) C6H5N2Cl + C6H5NH2 OH─

[2]

15. Give reason for the following:a) Aniline cannot be prepared by Gabriel phthalimide synthesis.b) An aqueous solution of methyl amine reacts with FeCl3 to form brown precipitate of hydrated ferric oxide.

[2]

16. State the pharmacological effect of the following drugs on human body.a) Dimetappb) Equanil

[2]

17. Explain the amphoteric behaviour of amino acids in aqueous solution. [2]

18. The standard free energy of formation of Cr2O3 is ─ 540 kJmol─1 and that of Al2O3 is ─ 827 kJmol─1. Can aluminium be used as a reducing agent to obtain chromium from Cr2O3?

[2]

19. a) Non-stoichiometric Cu(I) oxide is found to have the of composition Cu1.96O Account for the fact that this substance is a p-type semiconductor. b) A metal ‘M’ crystallizes in a cubic close packed structure. Its metallic radius is 125 pm.i) Determine the length of the side of the unit cell.ii) How many unit cells are there in 1.00 cm3 of the metal?

[3]

20. For the reaction :H2(g) + 2 AgCl(s) → 2 Ag (s) + 2H+ (0.1M) + 2 Cl─(0.1M)(1 atm)Go = ─ 42927 J at 25oC. Calculate the emf of the cell in which the given reaction takes place.

[3]

21. Nylons are named by giving numbers after the word nylon. The first number represents the number of carbon atoms in the diamine monomer and the second number represents the number of carbon atoms in the dicarboxylic acid monomer.a) Give the structural formulae of the monomers that make up nylon-6,10.b) Draw a part of the polymer chain of nylon-6,10.c) What intermolecular forces operate between neighbouring polymer chains?

[3]

22. A first order reaction is 50% complete in 30 min at 27oC and in 10 min at 47oC. Calculate the reaction rate constant at 27oC and the energy of activation for the reaction. R = 8.314 JK─1mol─1

[3]

23. Discuss the application of colloids in: [3]

of 5

a) Purification of drinking waterb) Tanning of leatherc) Obtaining rubber from latex.

24. a) Write balanced chemical equation for disproportionation reaction of H3PO3.b) What inspired N. Bartlett to carry out the reaction of xenon with PtF6?c) Oxygen exists as diatomic molecule whereas sulphur exists as S8 molecule. Why? OR

Explain the trends in the following properties of the hydrides of group 16.a) Acidic characterb) Boiling pointc) Thermal stability

[3]

25. Give reasons for the following:a) The variation in oxidation state of transition elements is by unity while for non-transition elements it is by 2.b) Eo (M3+/M2+) values of Sc and Fe are low but that of Zn and Mn are high.c) Transition metals form a large no. of complex compounds.

[3]

26. Cis – platin [Pt(NH3)2Cl2] is an anti-cancer agent because it inhibits the growth of tumours. It is found to be diamagnetic in nature.a) Predict the state of hybridization of the central metal atom / ion of the molecule.b) Draw the structure of the compound and that of its possible isomer(s).c) Write the IUPAC name cis – platin.

[3]

27. a) Explain with resonance structures why phenol on bromination with Br2 in CS2 does not produce m-Bromophenol but its ortho- and para- isomers.b) Write equation for the action of heated copper on vapours of tertiary – butyl alcohol.

[3]

28. a) Two elements P and Q form compounds of formulae PQ2 and PQ4. When dissolved in 20 g of benzene, 1.0 g of PQ2 lowers the freezing point by 2.3oC whereas 1.0 g of PQ4 lowers it by 1.3oC. Determine the atomic masses of P and Q. The molal depression constant of benzene is 5.1 K/m.b) Explain with the help of a graph the lowering of freezing point of water when glucose is dissolved in it. ORa) The vapour pressure of two liquids X and Y are 450 and 700 mm Hg respectively at 350 K. Calculate the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the mixture in the vapour phase.

b) Why does a solution of ethanol and water disobey Raoults law? Compare

[5]

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the boiling point of the solution to that of its pure components.

29. a) Account for the following : i) Solid PCl5 exists as an ionic solid. ii) SF6 is resistant to hydrolysis iii) Phosphine has lower boiling point than ammonia.b) Draw the spatial structures of the following molecules: i) Hypophosphorous acid ii) Pyrosulphuric acid ORa) Give a chemical equation in each case to support the following statements: i) Ammonia behaves as a Lewis base. ii) Concentrated sulphuric acid is a dehydrating agent. iii) Chlorine is a strong bleaching agent.b) Draw the spatial structures of molecules of: i) Iodine pentafluoride ii) Xenon hexafluoride

[5]

30. a) Give simple chemical tests to distinguish between the following pairs of compounds: i) Ethanol and propanone ii) Benzaldehyde and benzoic acidb) How can the following conversions be carried out? i) Benzonitrile to acetophenone ii) An alkyl halide into a primary amine whose molecule has one more carbon than the used alkyl halide molecule iii) acetic acid into acetaldehyde OR An organic liquid (A) of molecular formula C6H12O2 is hydrolysed by dilute NaOH solution to give compounds (B) and (C). (B) on treatment with dilute HCl gives (D). (C) on oxidation with acidified potassium dichromate also gives (D). (D) on strong heating with ammonia gives (E) which on treatment with bromine and NaOH forms compound (F) of composition C2H7N . (F) when heated with chloroform and ethanolic KOH produces an offensive odour. Write the structures of compounds (A) to (F). Write the sequence of reactions involved.

[5]

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OUR OWN HIGH SCHOOL, AL WARQA’A

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REHEARSAL EXAMINATION DEC.2009

MARKING SCHEME ─ CHEMISTRY ( 03.01.2009)

GRADE ─ XII M. Marks : 70

S.No. Value Points Point wise marks

Total marks

1. So that the surface of the electrode becomes suitable for the adsorption of hydrogen gas.

1 1

2. Due to honey- combe structure reactants of appropriate size get adsorbed on them easily.

1 1

3. It acts as a flux and. It lowers the melting point of alumina and improves its conductivity.

1 1

4. Cr2O72─ + 3 Sn2+ + 14 H+ → 2 Cr3+ + 3 Sn4+ + 7 H2O 1 1

5. This is because the number of current carrying ions per unit volume decreases on dilution.

1 1

6. 2─Cyclohexyl─4─methoxypentan─2─ol 17. DNA is the reserve of genetic information. It also carries

the code for protein synthesis.Various RNA molecules actually carry out protein synthesis.

½

½

1

8. Benzene─1,4─dicarbaldehyde 19. CdCl2

Each Cd2+ replaces 2 Ag+ from the crystal to maintain electrical neutrality. One of the Ag+ site is occupied by Cd2+ ion, the other remains vacant thereby creating Schottky defect.

½ ½

1

2

10. a) Order with respect to B = 0 and order with respect to A is 1.Rate = k[A][B]0

½ , ½

1

2

11. 58 and 6358 - +3 and +463 - +2 and +3 OR3d3 - V2+ , Cr3+

3d6 - Fe2+ , Co3+

11

11

2

12. NCERt text page 289 and 307 2

13. a) H2SO4 converts KI to HI and then oxidizes it to I2

b) It is a 1o halide hence reacts faster.11

2

14. a) C6H5N2Cl + CH3CH2OH C6H6 + CH3CHO + N2 + HClb) C6H5N2Cl + C6H5NH2 OH─

1 2

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C6H5N=NC6H4─NH2-p + HCl 1

15. a) C─X bond of C6H5─X is stabilized by resonance and difficult to break.b) Methyl amine is basic . It yields OH─ ions in solution which precipitates Fe3+ ions as Fe(OH)3.

1

1

2

16. a) Antihistamine- It interferes with natural action of histamine by competing with histamine for binding sites of receptor where histamine exerts its effect.b) Tranquilizer- It is used in controlling depression and hypertension.

½ + ½

½ + ½

2

17. Amino acids have both an acid group and a basic group. In aqueous solution basic ─NH2 group accepts proton and acidic ─ COOH group donates H+ ion resulting in the formation of zwitter ion.

1

1

2

18. 2 Cr + 3/2 O2 → Cr2O3 fGo = ─ 540 kJmol─1

2 Al + 3/2 O2 → Al2O3 fGo = ─ 827 kJmol─1

Required reaction: 2 Al + Cr2O3 → Al2O3 + 2 Cr

Go(total) = ─ 827 + 540 = ─ 287 kJmol─1

Since Go(total) for the coupled reaction is negative Al can be used as the reducing agent.

1

1

2

19. a) It shows metal deficiency defect. Some Cu+ ions are missing from the lattice. A corresponding no. of Cu+ ions are oxidized to Cu2+ to maintain electrical neutrality. Missing cations create positive holes. Hence p- type semiconductor.b) For fcc unit cell : a = 2√2r = 2 x 1.414 x 125 = 353.5 pm Volume of unit cell a3 = (353.5 x 10─10) cm3

= 4.42 x 10─23 cm3

No. of unit cells in 1 cm3 = 1/ 4.42 x 10─23

= 2.26 x 1022

1

½

½

1

3

20. Go = ─ nFEocellHence Eocell = ─ G o = 42927 = 0.2224 V nF 2 x 96500

Ecell = Eocell ─ 0.059 log[H + ] 2 [Cl ─ ] 2

n [H2] n = 2 = 0.2224 ─ 0.059 log [0.1] 2 [0.1] 2 2 [1]

= 0.2224 ─ 0.0295 log 10─4

½ ½

1

3

of 5

= 0.2224 + 0.118 = 0.3404 V

1

21. a) H2N-(CH2)6-NH2 and HOOC-(CH2)8-COOHb) [-HN-(CH2)6-NH-CO-(CH2)8-CO-]n

c) Intermolecular hydrogen bonds between CO- and –NH of neighbouring chains.

111

3

22. k = 0.693 / t ½ = 0.693 / 30 = 0.0231 min─1

log (k2 / k1) = [Ea/ 2.303 R] [1/ T1 ─ 1/T2]Rate at 27oC is 3 times the rate at 47oClog 3 x 2.303 x 8.314 x 300 x 320 320 ─ 300= 0.4771 x 2.303 x 8.314 x 300 x 320 20= 43848.5 J / mol= 43.8485 kJ / mol

½ ½½ ½

1

3

23. NCERT Text pages 144 - 145 1 x 3 324. a) 4 H3PO3 → 3 H3PO4 + PH3

b) Nearly the same first I.E. of O2 molecule and Xe and that O2PtF6 could be prepared.c) Oxygen atom being small in size can form pπ – pπ double bond while the bigger S atom cannot. ORa) Increases down the group from H2O to H2Te due to decrease in H─E bond dissociation enthalpy.b) Increases down the group due to increase in size of molecules. However water has the highest because of intermolecular H ─ bonding.c) Decreases down the group due to decrease in H─E bond dissociation enthalpy.

11

1

1

1

1

3

25. a) In transition elements variability in O.S arises due to incompletely filled or partially filled d-orbitals, while in non-transition elements it is due to inert pair effect and excitation of electrons.

b) Low values indicate stability of +3 more than +2 while high values indicate +2 state is more stable than +3.Sc : +3 state is stable because it has a noble gas configuration of Ar.Fe3+ is more stable due to half-filled 3d5 configuration. Zn2+ is more stable due to 3d10 configuration. Mn2+ is stable because of 3d5 configuration.

c) Due to small size of metal ions, their high ionic charges and availability of empty d-orbitals for bond formation.

1

1

1

3

26. a) dsp2 1 3

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b) cis – and trans- isomers

Cl NH3 Cl NH3

Pt Pt Cl NH3 H3N Cl

c) cis – Diamminedichloridoplatinum(II)

½ , ½

127. a) Resonance structures

Explanationb) CH3 CH3

│ heated Cu │ CH3─C─ OH CH3─C = CH2

│ CH3

11

1

3

28. a) Molar mass of PQ2 = 1000 x kf x w(solute) w(solvent) x Tf

= 1000 x 5.1 x 1.0 20 x 2.3 = 110.87 g / mol

Molar mass of PQ4 = 1000 x 5.1 x 1.0 20 x 1.3 = 196.15 g / mol

Molar mass of PQ4 ─ Molar mass of PQ2 = 2 x atomic mass of Q 196.15 ─ 110.87 = 85.28Hence atomic mass of Q = 42.64 uMolar mass of PQ2 = Atomic mass of P + 2 x Atomic mass of Q 110.87 = Atomic mass of P + 85.28Hence Atomic mass of P = 110.87 ─ 85.28 = 25.59 u

b) Graph and explanation – NCERT Text page 50 ORa) p(total) = po

1x1 + po2x2

600 = 450( x1) + 700 ( 1 ─ x1) 600 = 450( x1) + 700 ─ 700( x1)700( x1) ─ 450( x1) = 700 ─ 600 250( x1) = 100Hence x1 = 100/250 = 0.4 Mole fraction of X in liquid phase = 0.4Mole fraction of Y in liquid phase = 0.6 Partial pressure of X = 0.4 x 450 = 180 mm Hg Partial pressure of Y = 0.6 x 700 = 420 mm Hg

½

½

½

½

½

½

1,1

½

½ ½ ½

5

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Mole fraction of X in vapour phase = pX / pTotal = 180 / 600 = 0.3Mole fraction of Y in vapour phase = 0.7b) Because A─B forces are weaker than A─A and B─B forces. Hence positive deviation.Boiling point of solution will be lower than that of the pure components.

½ ½

1

1 29. a) i) Due to the presence of [PCl4]+ and [PCl6]─ ions .

ii) S is sterically protected by 6 F atoms. iii) Ammonia molecules are associated by intermolecular H─bonding.b) H3PO2 H2S2O7

ORa) Each equationb) IF5 XeF6

111

1,1

1 x 31, 1

5

30. a) i) Ethanol on esterification with ethanoic acid gives pleasant fruity smell of ester OR propanone gives red colour with 2,4-DNP ii) Benzaldehyde gives silver mirror with Tollens reagent or red ppt with Fehlings solutionBenzoic acid gives brisk effervescence with NaHCO3

b) i) C6H5CN + CH3MgBr C6H5C=NMgBr │ CH3

C6H5CCH3 H3O+

║ O

ii) R─X KCN RCN LiAlH4 RCH2NH2

H2O PCl5 H2

1

1

1

1

5

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iii) CH3COOH CH3COCl Pd- BaSO4

CH3CHO OR

[A] – CH3CH2COOCH2CH2CH3

[B] - CH3CH2COONa

[C] - CH3CH2CH2OH

[D] - CH3CH2COOH

[E] - CH3CH2CONH2

[F] - CH3CH2NH2

NaOH CH3CH2COONaCH3CH2COOCH2CH2CH3 + [A] CH3CH2CH2OH

CH3CH2COONa dil HCl CH3CH2COOH [B] [D]

K2Cr2O7 / H+

CH3CH2CH2OH CH3CH2COOH [C] NH3 , CH3CH2COOH CH3CH2CONH2

[E] CH3CH2NH2 NaOH / Br2

[F]

1

½ x 6

2

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